Problem: Find $\lim_{h\to 0}\dfrac{2\sqrt[3]{8+h}-2\sqrt[3]{8}}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac1{12}$ (Choice B) B $\dfrac{1}{6}$ (Choice C) C $4$ (Choice D) D The limit doesn't exist
Explanation: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $2\sqrt[3]{8+h}-2\sqrt[3]{8}$, we can tell that the function is $f(x)=2\sqrt[3]{x}$ and the $x$ -value is $8$. In other words, the limit expression is equal to $f'(8)$ for $f(x)=2\sqrt[3]{x}$. Let's find $f'(x)$ : $f'(x)=2\cdot \dfrac13\cdot {x^{^{{-\scriptsize\dfrac23}}}}=\dfrac{2}{3\sqrt[3]{x^2}}$ Now let's evaluate $f'(8)$ : $f'(8)=\dfrac{2}{3\sqrt[3]{8^2}}=\dfrac{1}{6}$ In conclusion, $\lim_{h\to 0}\dfrac{2\sqrt[3]{8+h}-2\sqrt[3]{8}}{h}=\dfrac16$.